---
title: "Convolve {1,2,1,−1} and {1,2,3,1}"
subject: "Signals and Systems"
chapter: 6
key: "semester_4/signals_and_systems:convolution-origin-given"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-origin-given
site: BCT Engineering
---

# Convolve {1,2,1,−1} and {1,2,3,1}

Signals and Systems · Chapter 6: Discrete-Time Systems

- Asked in: 2076 Bhadra, 6 marks; 2075 Baisakh, 5 marks (asked 2 times)
- Video: [Method to Find Discrete Convolution](https://www.youtube.com/watch?v=Nj8no2GMm8s) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-origin-given
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6

## Question

Perform convolution between signals: x₁[n] = {1, 2, 1, −1} with origin at second position from left and x₂[n] = {1, 2, 3, 1} with origin at first position from left.

## Short answer

With x₁[-1]=1,x₁[0]=2 and x₂[0]=1, tabular convolution yields y[n]={1,4,8,8,3,-2,-1} with y[0]=4, for -1≤n≤5.

## Answer

**Given (origin marked):**
- $x_1[n] = \{1, \underset{\uparrow}{2}, 1, -1\}$, so $x_1[-1] = 1$, $x_1[0] = 2$, $x_1[1] = 1$, $x_1[2] = -1$ ($-1 \le n \le 2$).
- $x_2[n] = \{\underset{\uparrow}{1}, 2, 3, 1\}$, so $0 \le n \le 3$.

$$
y[n] = x_1[n]*x_2[n] = \sum_k x_1[k]\,x_2[n-k]
$$

**Output range:** start $= -1 + 0 = -1$; length $= 4 + 4 - 1 = 7$; so $-1 \le n \le 5$.

#### Tabular method

Each row is one sample of $x_1$ times $x_2$ shifted to that sample's position:

| Row | $n=-1$ | $n=0$ | $n=1$ | $n=2$ | $n=3$ | $n=4$ | $n=5$ |
|---|---|---|---|---|---|---|---|
| $x_1[-1]\,x_2[n+1]$ | 1 | 2 | 3 | 1 |  |  |  |
| $x_1[0]\,x_2[n]$ |  | 2 | 4 | 6 | 2 |  |  |
| $x_1[1]\,x_2[n-1]$ |  |  | 1 | 2 | 3 | 1 |  |
| $x_1[2]\,x_2[n-2]$ |  |  |  | -1 | -2 | -3 | -1 |
| $y[n]$ | **1** | **4** | **8** | **8** | **3** | **-2** | **-1** |

#### Direct check of some values

- $y[-1] = x_1[-1]x_2[0] = 1$
- $y[0] = x_1[-1]x_2[1] + x_1[0]x_2[0] = 2 + 2 = 4$
- $y[1] = 1(3) + 2(2) + 1(1) = 8$
- $y[2] = 1(1) + 2(3) + 1(2) + (-1)(1) = 8$
- $y[3] = 2(1) + 1(3) + (-1)(2) = 3$
- $y[4] = 1(1) + (-1)(3) = -2$
- $y[5] = (-1)(1) = -1$

**Sum check:** $(\sum x_1)(\sum x_2) = 3 \times 7 = 21$, and $1 + 4 + 8 + 8 + 3 - 2 - 1 = 21$.

**Answer:**

$$
y[n] = \{1,\ \underset{\uparrow}{4},\ 8,\ 8,\ 3,\ -2,\ -1\}
$$

for $-1 \le n \le 5$ (the arrow marks $n = 0$, so $y[0] = 4$).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. x1 = {1, 2, 1, -1} with origin at the second position. Which sample positions does x1 occupy?
   A. 0 to 3
   B. -1 to 2
   C. -2 to 1
   D. 1 to 4
   Answer: B. The 2 is at n = 0, so the first sample 1 is at n = -1, and the last at n = 2. (Key point: Sample positions)

2. x2 occupies 0 to 3. Over which range does y = x1 * x2 run?
   A. -1 to 5
   B. 0 to 6
   C. -1 to 6
   D. -1 to 2
   Answer: A. Start = -1 + 0 = -1, length = 4 + 4 - 1 = 7, so the end is 5. (Key point: Output range)

3. What is y[0]?
   A. 2
   B. 4
   C. 8
   D. 1
   Answer: B. y[0] = x1[-1]x2[1] + x1[0]x2[0] = 1(2) + 2(1) = 4. (Key point: Value at origin)

4. What should the sum of all y[n] equal?
   A. 10
   B. 21
   C. 3 + 7 = 10
   D. 4
   Answer: B. (sum x1)(sum x2) = 3 x 7 = 21, matching 1+4+8+8+3-2-1. (Key point: Sum check)

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